Max/min on a continuous interval.

**cgiulz** · October 24th 2009, 01:01 PM

Suppose that $f(x)$ is continuous and $> 0$ on $I = \mathbb{R},$ and $\displaystyle\lim_{x\to \pm\infty}f(x) = 0.$

$(a)$ Prove $f(x)$ has no minimum on $I.$
$(b)$ Prove $f(x)$ has a maximum on $I$ (note that $I$ is not compact).
$(c)$ Prove $(b)$ under weaker hypothesis than positivity on all of $I$ .

**Jose27** · October 24th 2009, 01:33 PM

Originally Posted by cgiulz

Suppose that $f(x)$ is continuous and $> 0$ on $I = \mathbb{R},$ and $\displaystyle\lim_{x\to \pm\infty}f(x) = 0.$

$(a)$ Prove $f(x)$ has no minimum on $I.$
$(b)$ Prove $f(x)$ has a maximum on $I$ (note that $I$ is not compact).
$(c)$ Prove $(b)$ under weaker hypothesis than positivity on all of $I$ .

For (a) I'm assuming you mean a global minimum (otherwise this is not true) and it's obvious since that minimum would have to be $0$ contradicting the fact that $f>0$ .

For (c) we're going to assume $f \geq 0$ for some $x \in \mathbb{R}$ . Take an interval of the form $[-K,K]$ and take the maximum of f there say $f(x_0)=y_0>0$ (if it's zero we'll deal with it later). Now since $f(x) \rightarrow 0$ when $x \rightarrow \pm \infty$ we have two constants $K_1,K_2>0$ such that $\vert f(x) \vert < y_0$ whenever $x>K_1$ or $x<-K_2$ take $m= \max \{K_1,K_2 \}$ and so we have that $f$ attains a maximum $y$ in $[-m,m]$ with $y \geq y_0$ and since out of this interval the function never exceeds $y_0$ , $y$ is a global maximum.

If $y_0$ were $0$ and $f \leq 0$ in I we're finished. If it's $f>0$ on an unbounded interval apply a similar argument to the above, if $f>0$ on a bounded interval it's clear it has a maximum. And this covers all cases.

**cgiulz** · October 25th 2009, 05:48 PM

Can anyone show me where to start in order to prove the existence of a max by positivity?

Thanks.

[Forget it, I realized I can just pull it out of part (c)]

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